[In this post I use calculus that I don’t understand rigorously. I hope that is excused for this seemed a pretty interesting thing to discover.]
One is familiar with the Euler-Lagrange equation in one dimension, where the function that extremises necessiates the following equation on be satisfied:
Using (1) one can find as a differential equation, which can be completely solved if initial conditions are given. As an aside of some interest (in the Lagrangian formulation of classical mechanics), one can note that there can be at most two undetermined constants in the solution of (1), so taking and taking as known will provide a unique solution to , as will taking and . This equivalence of two initial conditions means that the behaviour of the function is completely specified for all defined if the initial coordinate and the derivative of the coordinate are given, and an equation of the form (1) is known; for that (1) can be derived by fixing the coordinates of two end-points (but not the derivatives of the coordinates) and imposing the conditions of extremum for the phase space of the functional lying between and .
This post aims to derive the two dimensional analogue of (1), i.e. what equation a function of two independent variables and , , must satisfy so that an integral of a functional,
is extremised. is a simply connected subset of , i.e. a patch colloquially, with a defined closed boundary (the boundary can be piecewise continuous, the only requirement is that it has only a finite number of points of non-differentiability). is assumed to be differentiable upto at least the second order, and so is . It is to be noted that need not be convex in general. But we have the property that any surface can be subdivided into a number of surfaces such that for each surface at no point the inverse tangent of slope (wrt to a cartesian coordinate system), if exists, shall not change counterclockwise and cross the or barriers. This will ensure that we can define (3) below in terms of complete integrals. Since any subsurface of must separately satisfy the extremisation principle, in this proof it shall be assumed that has the property above.
Since is differentiable upto the second order, and is of the property above, the form (2) can be written in a cartesian form using double integrals as
The notation I have used for the limits of the integrals in (3) is: and are the lower and upper limits of a small rectangular strip of of width at a specific , and likewise for interchanging and ; are the x-bounds of itself, and likewise are the y-bounds of . It is to be noted that all the limits of integration in (3) are boundary points.
Attempting to tackle the problem like the one-dimensional case, let us assume for where is the closed boundary of and is known; is then specified at the boundary of .
Assume that the specific function that extremises (1) is , i.e. is a local extremum for all within a sufficiently small neighbourhood of (of an infinite dimensional function space). Working analogously with this loose intuitive idea of an infinite dimensional function space, let us effect a variation in wrt , i.e. replace by a point close to , analogous to the general procedure we use for the one dimensional problem. This variation can be intuitively pictured as having a magnitude and a direction, and so we can write
where is the magnitude of the displacement along the drection of . Due to the boundary conditions being fixed, we have
Differentiating (4) wrt and and then wrt , one can get the equations
One can now say that for to extremise , at .
This means, from (2),
Expanding using the chain rule, one has
Using (7) the above can be compacted as
Consider the integral . It can be equated thus, using (3), and the fact that at boundary points:
Similarly, interchanging the roles of and as in (9) and again using (3), one can get
Using (9) and (10), (8) becomes
Since (11) must hold true if we took any subdomain of , we hence conclude that the term in crochets will always be identically 0. This gives us the required equation must satisfy to extremise (2) which we set out to find:
This holds true for all kinds of simply connected surfaces, convex or concave.
Application to the stretched membrane problem:
[Some calculation credit is due to more analytically adept people.]
We are given a closed frame in space, that forms a closed loop . In a three dimensional cartesian system the loop can be described by its height at each point from the x-y plane, and the x- and y-coordinates of the projection of the loop on the x-y plane. The loop is assumed to be without any kinks, i.e. if parameterised, all functions will be differentiable upto the second order. An elastic rubber membrane is stretched over the loop so that the edges of the membrane adhere to the loop. What will be the equation of the membrane?
A physical assumption is that the membrane tries to assume a shape that renders it with the minimal possible area, to assume the least potential energy position. This means that an integral of the form
is to be minimised (there is clearly no maximum limit to the area). Here is the projection of the loop on the x-y plane, and is a function which encodes how ‘steep’ the slope is at a particular point , i.e. if a small patch of the manifold of measure made an angle of with a plane parallel to the x-y plane, then .
One can obtain through some geometry (skipping the details) that . Plugging this form into equation (12) gives
Simplyfing the above one gets the simple and elegant (and almost Laplacian)
which has as initial values the coordinates of the loop. I’m not sure on how to show that that is sufficient to determine the membrane uniquely, if at all.